f(x) = 3x - 7 and g(x) = -2x - 6. Find (f o g)(4) Please help!?
First plug in g(x)=-2x-6 in f(x)=3x-7 in place of x :)
So would it be f(x) = 3(-2x - 6) - 7 ?
Yes f(g(x))=3(-2x-6)-7 Now put 4 in place of x and get (fog)(4)=.. or f(g(4))=..
Okay, so would it be f(g(4)) = -49 ?
Yeah right :)
Okay Thank you! could you help me with one more please ?
Sure :)
Okay, Rewrite \[\sqrt[3]{27x - 81} - 5\] to make it easy to graph using a translation. Describe the graph
I know that you would put 1/3 as an exponent but I don't know where
Well see this. \[\huge\ \sqrt[3]{a}=(a)^{1/3}\] Another example, \[\huge\ \sqrt[3]{x-2}=(x-2)^{1/3}\]
Okay so it would be (27x - 81)^1/3 - 5, right ?
Yes we can work more on it if you want :)
Yes please! Would this be the equation to make it easier to graph or is there more ?
Yes lets continue then take the 27 common. \[\huge\ [27(x-1)]^{1/3}-5 \] \[\huge\ [3^3 (x-1)]^{1/3}-5\]
Now you can cancel. \[\huge\ (x^3)^{(1/3)}=x\]
Okay, is this what the equation is or is this just stating what x equals ?
I am showing you how you can cancel.
The above statement was an example.
Oh okay
\[\huge\ [3^3 (x-1)]^{1/3}-5\] Can you further simplify this ?
Wait, I am not sure how you did the (x^3) ^1/3 = x
Would it be (9x - 1)^1/3 - 5 ? @hba
No.
I'm confused :/
\[\huge\ (3^3)^{1/3} ((x-1))^{1/3}-5\]Now can you do it ?
Lets see... (3^3)^1/3 would become just 3 correct?
Yeah :)
Okay so it would be 3(x - 1) - 5 ? or would I combine 3 and x - 1?
No. 3 (x-1)^{1/3}-5
Oh, I forgot the 1/3 sorry
Then what would I do?
Have some ice-cream :) Can i get some cookies ? I am the cookie monster.
C is for cookie and cookie is for me. P.S. I am out of cookies :'(
Wait, is that the answer ?
Yeah LoL
Oh haha, then I would just graph it and describe what it looks like right ?
I've got some cookies lol (:
cookies LoL :P Yeah you can graph it :) Now it is easy i guess :)
Okay lol, Thank you for your help!
Welcome Welcome :)
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